Name: Math 243: Calculus C
Exam 3 Solution Guide
26 April 2002

Always the beautiful answer who asks a more beautiful question.
-E. E. Cummings

Instructions: Show all work to receive full or partial credit. You may use a scientific calculator on this exam. All University rules and guidelines for student conduct are applicable.

1.

[20 pts] Reverse the order of integration of the following double integral.

$$\int_0^1 \int_x^{2x} f(x,y) dy dx.$$

The first step is to sketch the region of integration.

If we are going to integrate in y first, we must break the integral into two parts.

$$\int_0^1\int_{\frac{y}{2}}^y f(x,y) dx dy + \int_1^2\int_{\frac{y}{2}}^1 f(x,y) dx dy.$$

2.

Consider the solid object bounded by the surfaces z=0, z+y²=1, x+z=1 and -x+z=1.

(a)

[5 pts] Sketch the object labeling all axes carefully.

The toughest part of problems involving double and triple integrals is determining the proper region. After doodling around a little and roughly drawing each surface, you should see the intersection start to emerge. In the end, it will look like

which is not nearly as good as what I can do by hand. Or, if I ask Maple to sketch it,

(b)

[15 pts] If the density of the object is

f(x,y,z) = 1-z,

find the mass of the object.

Integrating the density over this region, we find that

$$\begin{eqnarray*}m & = & \int_{-1}^1 \int_0^{1-y^2} \int_{z-1}^{1-z} (1-z) dx dz... ... & = & \int_{-1}^1 \frac{2}{3} (y^6-1) dy \\ & = & \frac{8}{7} \end{eqnarray*}$$

This is one way to set up the triple integral. There are others, but this one is very simple.

3.

[20 pts] Find the volume of the hour-glass.

By symmetry, we can find the volume of one cone and then multiply by two. We see that the edge of the cone satisfies the equation z=2rif we use cylindrical coordinates. Thus, the total volume is

$$V = 2 \int_0^{2\pi} \int_0^1 \int_{2r}^2 r dz dr d\theta = \frac{4 \pi}{3}.$$

4.

[20 pts] Find the volume of the intersection of the cylinder $x^2+y^2 \leq 1$and the sphere $x^2+y^2+z^2 \leq 4$.

Either cylindrical or spherical coordinates will do the trick fairly directly for this problem. Both approaches will involve two pieces. Here, I present a solution using spherical coordinates. Let R₁ be the wedges at the top and bottom. Let R₂ be the cylinders with the wedges removed.

Again, we can use symmetry to compute the top half and multiply by two. The boundary between R₁ and R₂ can be found to be $\varphi=\frac{\pi}{6}$. For R₁, the limits of integration are fairly routine.

$$V_1 = 2 \int_0^2 \int_0^{2\pi} \int_0^{\frac{\pi}{6}} \rho^2... ...rphi d\theta d\rho = \frac{16 \pi}{3} \left(2-\sqrt{3} \right)$$

The second volume is a little more subtle because the outer bound on $\rho$ corresponds to r=1 in cylindrical coordinates. In other words, $\rho \sin \varphi = 1$, so $\rho = \csc \varphi$ is the outside boundary. Thus, the volume can be expressed by the integral

$$V_2 = 2 \int_0^{2\pi} \int^{\frac{\pi}{2}}_{\frac{\pi}{6}} \... ...\int^{\frac{\pi}{2}}_{\frac{\pi}{6}} \csc^2 \varphi d \varphi.$$

At this point, some students stopped because they could not remember that the anti-derivative of $\csc^2 \varphi$ is $\cot \varphi$. This is understandable, and getting this far is worth most of the points for the problem. To complete the problem, we see that

$$V_2 = \left. \frac{4 \pi}{3} \cot \varphi \right \vert^{\frac{\pi}{2}}_{\frac{\pi}{6}} = \frac{4 \pi \sqrt{3}}{3}.$$

Adding the two together, we arrive at the solution,

$$V = V_1 + V_2 = \frac{4 \pi}{3} \left(8-3\sqrt{3} \right) = \frac{32 \pi}{3} - 4 \sqrt{3} \pi.$$

5.

[20 pts] Suppose f(x,y) = x² y. Find

$$\int_C f(x,y) ds$$

where C is the line segment running from (0,0) to (3,4).

We can parametrize this line as

$$\v r (s,t) = \langle 3,4 \rangle t.$$

Then, we find that ds = 5 dt. Now, it is simply a matter of computing the path integral:

$$\int_0^1 (3t)^2(4t) (5dt) = 45.$$